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What are two uses of total internal reflection? (GCSE Physics)

jeka94

Answer:

1) Used in optical instruments such as telescopes.

2) Used to form mirages

Explanation:

1) Perhaps the simplest example of this is the astronomical refractor telescope with a right-angle eyepiece holder. Astro scopes mostly point at things high in the sky, but it is inconvenient to place your eye low to look up through the scope, so the light path is bent 90 degrees just before the eyepiece. This can be done with a mirror, but using a simple 45 degree prism (internal angles 45, 45, and 90 degrees) will do the 90 degree bend more efficiently.

2) So when a light pass from cold air to hot air light tends to bend from its path which is known as refraction. As the light get refracted it reaches to a point where the light tends to form 90 degree angle.

3 0

2 years ago

Expresa en terminos de la unidad fundamental

Svet_ta [14]

Answer:

imma be honest I dint really know

5 0

2 years ago

A ball is thrown straight up from the edge of the roof of a building. A second ball is dropped from the roof 2.0 s later. You ma

malfutka [58]

Answer:

4.9 m/s

Hope this helps! c:

Explanation:

3 0

2 years ago

What is the net torque on the square plate, with sides 0.2 m, from each of the three forces? F1=18 N, F2=26 N, and F3=14 N. Use

marshall27 [118]

Answer:

The net torque on the square plate is 2.72 N-m.

Explanation:

Given that,

Side = 0.2 m

Force $F_{1}=18\ N$

Force $F_{2}=26\ N$

Force $F_{3}=14\ N$

We need to calculate the torque due to force F₁

Using formula of torque

$\tau_{1}=-F_{1}d_{1}$

$\tau_{1}=-F_{1}\times\dfrac{a}{2}$

Put the value into the formula

$\tau_{1}=-18\times\dfrac{0.2}{2}$

$\tau_{1}=-1.8\ N-m$

We need to calculate the torque due to force F₂

Using formula of torque

$\tau_{2}=F_{2}d_{2}$

$\tau_{2}=F_{2}\times\dfrac{a}{2}$

Put the value into the formula

$\tau_{2}=26\times\dfrac{0.2}{2}$

$\tau_{2}=2.6\ N-m$

We need to calculate the torque due to force F₃

Using formula of torque

$\tau_{3}=F_{3}d_{3}$

$\tau_{3}=(F_{3}\sin45+F_{3}\cos45)\times\dfrac{a}{2}$

Put the value into the formula

$\tau_{3}=0.1(14\sin45+14\cos45)$

$\tau_{3}=1.92\ N-m$

We need to calculate the net torque on the square plate

$\tau=\tau_{1}+\tau_{2}+\tau_{3}$

$\tau=-1.8+2.6+1.92$

$\tau=2.72\ N-m$

Hence, The net torque on the square plate is 2.72 N-m.

3 0

2 years ago

13) The condenser pressure is 417.4 psig for r-410A and the condenser outlet temperature is 108f. how much subcooling is there i

Tanzania [10]

Answer:

12°F

Explanation:

Calculation for how much subcooling is there in the condenser

Since the CONDENSING TEMPERATURE for 417.4 psig discharge pressure is 120 degrees (120°) which means that the amount of subcooling that is there in the condenser will be calculated using this formula

Amount of Condenser subcooling= Condensing Temperature discharge pressure -Condenser outlet temperature

Let plug in the formula

Amount of Condenser subcooling=120°-108f

Amount of Condenser subcooling=12°F

Therefore the amount of subcooling that is there in the condenser will be 12°F

8 0

2 years ago